Assume for answers \form is same type as \newcommand{\form}[1]{\emph{#1}}. But delay expansion until the end. Just in case.

Probably requires several nested \def cases taking two variables each and a \if to decide when to halt the loop ... like when allowing more than 9 arguments to a macro? Or use etoolbox for something better?

If separator words efg jkl are not the same from one place to the next, certain they they are of some specific length, assume efg jkl <=> always 9 characters separating one input from the next. While abc EDF xyz are of different lengths possibly, but always three separate words.

CLARIFICATION: The xyz can be arbitrary. But the sentence is periodic in that the 3rd, 8th, 13th, 18th word is \form wrapped. If it exists. At least 3 words exists. But can be 3, and if not 3 then 8, and if not 8 then 13, as long as needed, ends when no separator behind three words which can all differ ... The separator " efg jkl " does repeat with certainty however and can be caught to parse.

@GuidoJorg I'm not asking what \form should do, but to which parts it should be applied. Some real world examples might help to understand. As the question stands, it's impossible to guess.
– egregMar 22 at 22:41

@egreg to the 3rd, 8th, 13th, 18th ... word in the sentence. Separator can be assumed to repeat. \form affects only these parts. And sentence can be arbitrarily long and will end without the separator.
– Guido JorgMar 22 at 22:46

@GuidoJorg Why should \fun{{...}} give the same result as \fun{...}? I see no reason to.
– egregMar 23 at 14:06

By itself: A displays as {A}. In the use case, composition is desirable. If the text into \fun is generated by some macro, not directly entered, \fun won't process it. Because it seems the result is treated as {A} rather than A. Even though it is A. Single object. \fun sees no separators when Ab Cd Ef = \macro{test} and \fun{\macro{test}}.
– Guido JorgMar 23 at 14:13

(updated this answer signficantly to allow for multiple (expandable) macros in the argument of \fun.)

Here's a LuaLaTeX-based solution. It can handle multiple, expandable macros in the argument of \fun. The Lua code first splits the (expanded) input string into separate words, taking note of punctuation characters, if present. It then proceeds to print them, encasing the 3rd, 8th, 13th, 18th words in the \form macro. (Mathematically speaking, the selection criterion is that the word's position in the table, modulo 4, equals 3.) Non-ASCII UTF8-encoded characters are fine (because the unicode.utf8.gmatch function rather than the "basic" string.gmatch function is employed.)

If you need multi-layer parsing, say that efg jkl is the trigger to separate larger subgroups of arguments, then we have the following (note: efg jkl is not considered an argument, but an argument separator):

Can you say how to get a sentence in one line to occur, rather than a list? Of the rightmost list in the image? (Also for learning how to write such macros, how does listofitems pull this off, can you suggest? I assume it checks \ifx style whether the argument separator occurs or not to decide to loop and relies on a \def that that takes one argument and has two cases ... Or calls a package with check for \ifempty?)
– Guido JorgMar 22 at 22:36

@GuidoJorg As to your request, to make a sentence of the right hand column is not really to parse the list at all, but just to typeset the argument. So could you clarify what you are asking?
– Steven B. SegletesMar 22 at 22:40

The result looking for is what would be printed had abc1 EDF1 \form{xyz1} efg jkl abc2 EDF2 \form{xyz2} efg jkl abc3 EDF3 \form{xyz3} efg jkl abc4 EDF2 \form{xyz4} ... been entered manually. When in fact \fun{abc1 EDF1 xyz1 efg jkl abc2 EDF2 xyz2 efg jkl abc3 EDF3 xyz3 efg jkl abc4 EDF2 xyz4 ...} was entered manually. Though the knowledge of how to parse the list completely is useful to know, as is the package, didn't know about it :)
– Guido JorgMar 22 at 22:42